Question

Advanced photodiode detectors have a second light-emitting diode, operating at a wavelength of 2.0 × 10-7 m, to detect even smaller smoke particles from smoldering flames. What is the frequency difference between the two light beams?A) 12.0 × 10^15 HzB) 3.0 × 10^15 HzC) 2.0 × 10^15 HzD) 1.0 × 10^15 Hz

Answer 1

Answer:

Frequency, $f=1.5\times 10^{15}\ Hz$

Explanation:

It is given that,

Wavelength of the light- emitting diode, $\lambda=2\times 10^{-7}\ m$

We need to find the frequency difference between the two light beams. It can be calculated using the following relation as :

$c=f\times \lambda$

$f=\dfrac{c}{\lambda}$

$f=\dfrac{3\times 10^8}{2\times 10^{-7}}$

$f=1.5\times 10^{15}\ Hz$

So, the frequency difference between the two light beams is $1.5\times 10^{15}\ Hz$. Hence, this is the required solution.